Can the average value of a function be negative?

Yes. If the function spends more of its time below the x-axis than above it during the given interval, the definite integral will be negative, making the average value negative as well.

Does the function ever actually equal its average value?

Yes, as long as the function is continuous. The Mean Value Theorem for Integrals guarantees that a continuous function will hit its average value at least once somewhere inside that interval.

How is this different from the average rate of change?

The average rate of change tells you how fast the function is changing on average (the slope between the start and end points). The average value tells you the average height of the function itself over that interval.

What is the average value of a function?

The average value of a continuous function over an interval is a single number representing its typical height. You calculate it by taking the definite integral of the function over the interval and dividing by the interval's length.

Think of a glass of water that was just bumped. The surface is wavy, with high spots and low spots. If you wait for the water to settle, the surface becomes perfectly flat. That flat, settled water line is the average value of the original wavy surface.

The Formula

To find the average value of a function $f(x)$ on the interval $[a, b]$ , you use the formula: $f_{avg} = \frac{1}{b-a} \int_{a}^{b} f(x) dx$ . The integral part, $\int_{a}^{b} f(x) dx$ , calculates the total net area under the curve. Dividing by $(b-a)$ , which is the width of the interval, essentially 'levels out' that area into a rectangle. The height of that resulting rectangle is the average value.

Why it Works

If you want to find the average of five test scores, you add them all up and divide by five. A continuous function has an infinite number of points, so we cannot simply add them up and divide by infinity. Instead, the definite integral acts as our continuous version of 'adding everything up'. Dividing by the length of the interval, $b-a$ , is the continuous version of dividing by the number of items.

Where Students Slip

The most common mistake is forgetting to multiply by the fraction $\frac{1}{b-a}$ in front of the integral. Students will often correctly set up and evaluate the integral, finding the total area, but forget to divide by the width to find the average height. Another common trap is confusing 'average value' with 'average rate of change'. The average rate of change uses the algebra slope formula, while the average value uses an integral.

Worked through

Find the average value of $f(x) = x^2$ on the interval $[0, 3]$ .

First, identify $a = 0$ , $b = 3$ , and $f(x) = x^2$ . The width of our interval is $b - a = 3 - 0 = 3$ .

Set up the average value formula: $f_{avg} = \frac{1}{3-0} \int_{0}^{3} x^2 dx$ .

Next, find the antiderivative of $x^2$ , which is $\frac{x^3}{3}$ .

Now, evaluate the definite integral from 0 to 3: $\int_{0}^{3} x^2 dx = \frac{3^3}{3} - \frac{0^3}{3} = \frac{27}{3} - 0 = 9$ .

Finally, multiply by the $\frac{1}{3}$ out front: $f_{avg} = \frac{1}{3} (9) = 3$ .

The average value of $f(x) = x^2$ on $[0, 3]$ is 3.

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Where this comes from: OpenStax Calculus Volume 1, Chapter 1: Applications of Integration · Stewart Calculus, Chapter 6: Applications of Integration